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Suppose we have, $$y_t = a + {\alpha}y_{t-1}+u_t$$ for $t>k$, where $k$ is a positive integer and $\alpha \in (0,1)$. And, $$y_t = b + {\alpha}y_{t-1}+u_t$$ for $t\leq k$. And assume that $a$ and $b$ are two different real constants. $u_t$ are iid with mean $0$ and a variance, $\sigma$ constant and finite. $t \in \mathbb{Z}$.

Now, I am trying to get the moving average of the infinity order process of this. Normally, $\alpha$ being in the interval $(0,1)$, implies that ${sup}_t \ E(y_t)$ is finite. Thus, the space of sequences of $y_t$s is Banach space and we proceed with geometric series sum.

In this case, do we have to still show that ${sup}_t \ E(y_t)$ is finite? Or, $\alpha$ being in the interval $(0,1)$ implies this?

Thank you in advance.

user58697
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user722271
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1 Answers1

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Pick any $t\in\mathbb{Z}$. Then, it holds

$$E[y_t]\le \max\{|a|,|b|\}+\alpha E[y_{t-1}]$$

Supposing, there is an initial value $l\in \mathbb{Z}$, such that $E[y_l]=C<\infty$, you find

$$E[y_t]\le \max\{|a|,|b|\}\sum_{k=0}^{t+|l|-1}\alpha^k+ |E[y_{l}]| \sum_{k=0}^{t+|l|}\alpha^k \le \frac{1}{1-\alpha}(C+\max\{|a|,|b|\})<\infty$$

The constant does not depend on $t\in\mathbb{Z}$ and therefore $$\sup_{t\in\mathbb{Z}}E[y_t]<\infty$$

user408858
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